Non-Linear Simultaneous EquationsDate: 12/07/2003 at 15:41:14 From: Alroy Subject: solving a complex system of simultaneous equations Solve the simulataneous equations ab + c + d = 3 bc + d + a = 5 cd + a + b = 2 da + b + c = 6 where a, b, c, d are real numbers. I keep arriving at the point where everything cancels out and I am left with 0 = 0... Date: 12/07/2003 at 19:10:02 From: Doctor Schwa Subject: Re: solving a complex simultaneous equations Hi Alroy, This is a tough one! To me, it looks useful to combine the first and third equations, because they seem to have similar variables. I'll add them: ab + a + b + cd + c + d = 5. Now one of my favorite tricks comes into play. Note that ab + a + b is not a nicely factorable expression. But ab + a + b + 1 is -- it's just (a+1)(b+1). So, add 2 to both sides: ab + a + b + 1 + cd + c + d + 1 = 7 (a+1)(b+1) + (c+1)(d+1) = 7 Similarly (b+1)(c+1) + (a+1)(d+1) = 13 Adding these together gives (a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20 so ((a+1) + (c+1))*((b+1) + (d+1)) = 20 Now, if a, b, c, d are yucky numbers, this will still be a big pain. But, looking at the equations, it seems like you can guess and check some small integers at this point, integers that multiply to make 20... In fact, that might be a good approach to take trom the beginning, simply guessing and checking. Another possible idea of how to combine the questions is to take (Eq 1) + (Eq 3) - (Eq 2) - (Eq 4) so that the a+b+c+d subtracts out. Then you get ab + cd - bc - da = -6 (a-c)*(b-d) = -6 and again the guessing and checking is pretty easy if the numbers are small integers. Well, that's one style, and maybe it works well for you. Here's another: ab + c + d = 3 bc + d + a = 5 cd + a + b = 2 da + b + c = 6 Solve for a, either a = 5 - d - bc or a = 2 - b - cd take your pick. Substitute one of those into the first equation: (5-d-bc)*b + c + d = 3 Now solve something for b, but eliminate a first. For example, d*(5-d-bc) + b + c = 6 5d - d^2 -bdc + b + c = 6 b = (6 - 5d + d^2 - c)/(1 - cd) You can see that this will eventually solve for all the variables, but it will give you perhaps a quite ugly equation for d in the end. When you substitute this b into (5-d-bc)*b + c + d = 3 you'll get quite a messy-looking thing. So, try something else! Continuing from the first line of this attempt, where I started solving for a: 5 - d - bc = 2 - b - cd or 3 = (d-b) - c*(d-b) 3 = (1-c)*(d-b) But we could also get (d-b) by combining the first and fourth equations, eliminating c: 3 - ab - d = 6 - da - b or -3 = (1-a)*(d-b) Ah, that's the best approach so far, isn't it? We already know 1-a and 1-b are opposites at this point: that's the most concrete conclusion I've created in all these attempts. So (1-a) = -(1-b) 1-a = -1 + b a+b = 2 Among other things, that means right away that cd = 0, so c = 0 or d = 0, and checking those two cases should help you narrow things down fairly quickly. I hope one of these methods helps! Feel free to write back if you'd like some more hints. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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